OCR MEI FP1 2010 January — Question 5 6 marks

Exam BoardOCR MEI
ModuleFP1 (Further Pure Mathematics 1)
Year2010
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeStandard summation formulae application
DifficultyModerate -0.8 This is a straightforward application of standard summation formulae requiring expansion of the product, splitting into separate sums, and applying Σr and Σr² formulae. While it involves algebraic manipulation and is from Further Maths, it's a routine textbook exercise with no problem-solving insight needed—just mechanical application of known results.
Spec4.06a Summation formulae: sum of r, r^2, r^3
5 Use standard series formulae to show that \(\sum _ { r = 1 } ^ { n } ( r + 2 ) ( r - 3 ) = \frac { 1 } { 3 } n \left( n ^ { 2 } - 19 \right)\).
Question 5:
AnswerMarks Guidance
AnswerMark Guidance
\(\sum_{r=1}^{n}(r+2)(r-3) = \sum_{r=1}^{n}(r^2-r-6)\)
\(= \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - 6n\)M1 Separate into 3 sums
\(= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 6n\)A2 −1 each error
\(= \frac{1}{6}n\left[(n+1)(2n+1)-3(n+1)-36\right]\)M1 Valid attempt to factorise (with \(n\) as a factor)
\(= \frac{1}{6}n(2n^2-38) = \frac{1}{3}n(n^2-19)\)A1, A1 [6] Correct expression c.a.o.; Complete, convincing argument 
# Question 5:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{n}(r+2)(r-3) = \sum_{r=1}^{n}(r^2-r-6)$ | | |
| $= \sum_{r=1}^{n}r^2 - \sum_{r=1}^{n}r - 6n$ | M1 | Separate into 3 sums |
| $= \frac{1}{6}n(n+1)(2n+1) - \frac{1}{2}n(n+1) - 6n$ | A2 | −1 each error |
| $= \frac{1}{6}n\left[(n+1)(2n+1)-3(n+1)-36\right]$ | M1 | Valid attempt to factorise (with $n$ as a factor) |
| $= \frac{1}{6}n(2n^2-38) = \frac{1}{3}n(n^2-19)$ | A1, A1 **[6]** | Correct expression c.a.o.; Complete, convincing argument |

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5 Use standard series formulae to show that $\sum _ { r = 1 } ^ { n } ( r + 2 ) ( r - 3 ) = \frac { 1 } { 3 } n \left( n ^ { 2 } - 19 \right)$.

\hfill \mbox{\textit{OCR MEI FP1 2010 Q5 [6]}}
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Q1 5 Q2 7 Q3 6 Q4 6 Q5 6 Q6 6 Q7 12 Q8 12 Q9 12